3.67 \(\int \frac{\csc ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=105 \[ -\frac{\sqrt{b} (a-b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} f}-\frac{(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac{(a-b)^2 \cot (e+f x)}{a^3 f}-\frac{\cot ^5(e+f x)}{5 a f} \]

[Out]

-(((a - b)^2*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(7/2)*f)) - ((a - b)^2*Cot[e + f*x])/(a^3*f) -
((2*a - b)*Cot[e + f*x]^3)/(3*a^2*f) - Cot[e + f*x]^5/(5*a*f)

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Rubi [A]  time = 0.115417, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3663, 461, 205} \[ -\frac{\sqrt{b} (a-b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} f}-\frac{(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac{(a-b)^2 \cot (e+f x)}{a^3 f}-\frac{\cot ^5(e+f x)}{5 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

-(((a - b)^2*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(7/2)*f)) - ((a - b)^2*Cot[e + f*x])/(a^3*f) -
((2*a - b)*Cot[e + f*x]^3)/(3*a^2*f) - Cot[e + f*x]^5/(5*a*f)

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\csc ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a x^6}+\frac{2 a-b}{a^2 x^4}+\frac{(a-b)^2}{a^3 x^2}-\frac{(a-b)^2 b}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(a-b)^2 \cot (e+f x)}{a^3 f}-\frac{(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac{\cot ^5(e+f x)}{5 a f}-\frac{\left ((a-b)^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}\\ &=-\frac{(a-b)^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} f}-\frac{(a-b)^2 \cot (e+f x)}{a^3 f}-\frac{(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac{\cot ^5(e+f x)}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.79873, size = 103, normalized size = 0.98 \[ \frac{-\sqrt{a} \cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+a (4 a-5 b) \csc ^2(e+f x)-25 a b+15 b^2\right )-15 \sqrt{b} (a-b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{15 a^{7/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

(-15*(a - b)^2*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - Sqrt[a]*Cot[e + f*x]*(8*a^2 - 25*a*b + 15*b^2
+ a*(4*a - 5*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4))/(15*a^(7/2)*f)

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Maple [B]  time = 0.083, size = 191, normalized size = 1.8 \begin{align*} -{\frac{1}{5\,fa \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}-{\frac{2}{3\,fa \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{b}{3\,f{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{1}{fa\tan \left ( fx+e \right ) }}+2\,{\frac{b}{f{a}^{2}\tan \left ( fx+e \right ) }}-{\frac{{b}^{2}}{f{a}^{3}\tan \left ( fx+e \right ) }}-{\frac{b}{fa}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+2\,{\frac{{b}^{2}}{f{a}^{2}\sqrt{ab}}\arctan \left ({\frac{b\tan \left ( fx+e \right ) }{\sqrt{ab}}} \right ) }-{\frac{{b}^{3}}{f{a}^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x)

[Out]

-1/5/f/a/tan(f*x+e)^5-2/3/f/a/tan(f*x+e)^3+1/3/f/a^2/tan(f*x+e)^3*b-1/f/a/tan(f*x+e)+2/f/a^2/tan(f*x+e)*b-1/f/
a^3/tan(f*x+e)*b^2-1/f*b/a/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+2/f*b^2/a^2/(a*b)^(1/2)*arctan(b*tan(f
*x+e)/(a*b)^(1/2))-1/f*b^3/a^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.9664, size = 1319, normalized size = 12.56 \begin{align*} \left [-\frac{4 \,{\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 20 \,{\left (4 \, a^{2} - 11 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 60 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{60 \,{\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )}, -\frac{2 \,{\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \,{\left (4 \, a^{2} - 11 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 30 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{30 \,{\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/60*(4*(8*a^2 - 25*a*b + 15*b^2)*cos(f*x + e)^5 - 20*(4*a^2 - 11*a*b + 6*b^2)*cos(f*x + e)^3 - 15*((a^2 - 2
*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a*b + b^2)*sqrt(-b/a)*log(((a^2 +
6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 + 4*((a^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f*x + e
))*sqrt(-b/a)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2))*s
in(f*x + e) + 60*(a^2 - 2*a*b + b^2)*cos(f*x + e))/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)*si
n(f*x + e)), -1/30*(2*(8*a^2 - 25*a*b + 15*b^2)*cos(f*x + e)^5 - 10*(4*a^2 - 11*a*b + 6*b^2)*cos(f*x + e)^3 -
15*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a*b + b^2)*sqrt(b/a)*a
rctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*(a^2 - 2*a*b
+ b^2)*cos(f*x + e))/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)*sin(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6/(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.46882, size = 204, normalized size = 1.94 \begin{align*} -\frac{\frac{15 \,{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )}}{\sqrt{a b} a^{3}} + \frac{15 \, a^{2} \tan \left (f x + e\right )^{4} - 30 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 5 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/15*(15*(a^2*b - 2*a*b^2 + b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/(sq
rt(a*b)*a^3) + (15*a^2*tan(f*x + e)^4 - 30*a*b*tan(f*x + e)^4 + 15*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2
- 5*a*b*tan(f*x + e)^2 + 3*a^2)/(a^3*tan(f*x + e)^5))/f